Poynting Vector

Claimed by Tanner Shaw (tshaw30)

The Poynting vector represents the direction and magnitude of the flux in energy from an electromagnetic field. It was originally discovered by John Henry Poynting in 1884.

The Main Idea

The poynting vector describes the magnitude and direction of the flux in energy from an electromagnetic field. It is measured in units of Watts per meter squared ([math]\displaystyle{ \frac{W}{m^2}) }[/math]

A Mathematical Model

A graphical representation of the poynting vector

The poynting vector can be derived by the equation

[math]\displaystyle{ \vec{S} = {\frac{1}{μ_0}}\vec{E}\times\vec{B} }[/math]

where E is the electric field vector, and B is the magnetic field vector, and μ₀ is the magnetic constant. A cross product is needed to compute the poynting vector, so being familiar with the right-hand-rule can help to conceptualize and/or check your answer for problems asking to find it.

A Computational Model

The poynting vector can be used to predict the direction of energy movement in many different scenarios. Anything that produces electromagnetic waves can have the flux of energy produced from it mapped. One common example is an antenna broadcasting radio waves. A good demonstration of the potential for poynting vectors can be seen from Wolfram Alpha's Demonstration of Electromagnetic Waves from an Antenna. This demo requires the download of their Wolfram CDF player. In the demonstration, you can adjust variables such as the length of the antenna, peak current in the antenna, etc., and see how it affects poynting vectors in the space around the antenna.

Examples

Simple

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Simple Problem

An electron is travelling in the -X direction in the XY plane. At a certain instant, you observe the electron from a location as seen in the diagram. What is the direction of the poynting vector from your location?

The equation for finding the poynting vector is given as [math]\displaystyle{ \vec{S} = {\frac{1}{μ_0}}\vec{E}\times\vec{B} }[/math]. Therefore, a cross product of the electric field vector and the magnetic field vector at the observation location must be taken. This will require the concept of right-hand-rule. The electric field from the observation location points towards the electron, in the +X and +Y directions.

The magnetic field can be found by using the right-hand-rule of the velocity of the electron and the position vector. The electron is moving in the -X direction, and the position vector points in both the -X and -Y directions. Using right-hand-rule, the cross product of those two vectors will point in the +Z direction.

As the electric field vector points in the +X and +Y directions, and the magnetic field points in the +Z direction, using right-hand-rule, the poynting vector will point in the directions +X and -Y.

Moderate

Moderate Problem

A proton is moving in the +X direction at two thirds the speed of light in the XY plane. At a certain instance, your observation location is 3e-8 m in the -Y direction from the proton. Calculate the poynting vector at this observation location from the moving proton.

In order to solve this problem, we must calculate both the electric and magnetic fields caused by the moving particle at the observation location. To solve for the electric field we will use the formula for calculating electric fields from point charges, [math]\displaystyle{ \vec{E} = {\frac{1}{4πε_0}}{\frac{Q}{r^2}}\hat{r} }[/math].

Therefore, we must solve:

[math]\displaystyle{ \vec{E} = (9\times10^9\frac{N\times m^2}{C^2}){\frac{(1.6\times10^{-19} C)}{({3\times10^{-8} m})^2}} \lt 0,-1,0\gt }[/math]

Which will give us:

[math]\displaystyle{ \vec{E} = \lt 0, 1.6\times10^6,0\gt \frac{N}{C} }[/math]

Now that we have the electric field at the observation location, all we have left to calculate is the magnetic field. The equation of the magnetic field due to a moving point charge is [math]\displaystyle{ \vec{B} = {\frac{μ_0}{4π}} Q{\frac{\vec{v}\times\hat{r}}{r^2}} }[/math]

Therefore, we must solve:

[math]\displaystyle{ \vec{B} = (1\times10^{-7}{\frac{T\times m}{A}})(1.6\times10^{-19}C){\frac{\lt 2\times10^8,0,0\gt {\frac{m}{s}} \times\lt 0,-1,0\gt }{(3\times10^{-8}m)^2}} }[/math]

Which gives us:

[math]\displaystyle{ \vec{B} = \lt 0,0,-3.56\times10^{-3}\gt T }[/math]

Using the equation [math]\displaystyle{ \vec{S} = {\frac{1}{μ_0}}\vec{E}\times\vec{B} }[/math], we can find the answer.

[math]\displaystyle{ \vec{S} = (\frac{1}{4π\times10^{-7}})\lt 0, 1.6\times10^6,0\gt \frac{N}{C}\times\lt 0,0,-3.56\times10^{-3}\gt T }[/math]

[math]\displaystyle{ \vec{S} = \lt -4.53\times10^9, 0, 0\gt \frac{W}{m^2} }[/math]

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History

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John Henry Poynting

John Henry Poynting, an English physicist and professor, was the first to derive the equation that described the direction of electromagnetic flux. His work for the equation was published in 1884. Nikolay Umov and Oliver Heaviside both independently discovered the poynting vector as well.

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